Abstract. An element a of a commutative ring R is nilregular if and only if x is nilpotent whenever ax is nilpotent. More generally, an ideal I of R is nilregular if and only if x is nilpotent whenever ax is nilpotent for all a ∈ I. We give a direct proof that if R is Noetherian, then every nilregular ideal contains a nilregular element. In constructive mathematics, this proof can then be seen as an algorithm to produce nilregular elements of nilregular ideals whenever R is coherent, Noetherian, and discrete. As an application, we give a constructive proof of the Eisenbud-Evans-Storch theorem that every algebraic set in n-dimensional affine space is the intersection of n hypersurfaces.AMS Classification: 03F65 (14M10)
The nilregular element propertyLet R be a commutative ring with unit and N its nilradical, i.e. the ideal consisting of its nilpotent elements. We define an element a (respectively, an ideal I) of R to be nilregular if and only if x ∈ N whenever ax ∈ N (respectively, ax ∈ N for all a ∈ I). So an ideal I is nilregular precisely when the transporter ideal (N : I) = {x ∈ R : xI ⊆ N } is contained in N . We present a method to find nilregular elements of nilregular ideals when R is Noetherian. For this, we interpret first the property of being nilregular in a topological way.As usual, let D(a) be the set of prime ideals p of R such that a / ∈ p, and let D(a 1 , . . . , a n ) stand for the union of D(a 1 ), . . . , D(a n ). The intersection of D(a) and D(b) is D(ab), and D(a) is a subset of D(a 1 , . . . , a n ) if and only if a belongs to the radical of the ideal (a 1 , . . . , a n ) generated by a 1 , . . . , a n . In particular, D(a) = ∅ precisely when a ∈ N .It is well-known that the D(a) with a ∈ R form a basis of opens for the Zariski topology on the prime spectrum (the set of all prime ideals) of R. It follows that a ∈ R is nilregular if and only if D(a) is dense for the Zariski topology. Remark 1.2. D(a 1 , . . . , a n ) is dense if and only if (a 1 , . . . , a n ) is a nilregular ideal. Theorem 1.3. Let R be Noetherian, and a 1 , . . . , a n ∈ R. If D(a 1 , . . . , a n ) is dense, then the ideal (a 1 , . . . , a n ) contains a nilregular element.Proof. If D(x) = ∅, then there exists i such that D(xa i ) = ∅, because D(a 1 , . . . , a n ) is dense. Hence if the ring is nontrivial, then we can inductively build a sequence b 0 , b 1 , . . .